Problem 44 - Chemistry An Atoms First Approach

Question

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\). The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{K}\). What is the final volume of the balloon?

Step-by-Step Solution

Answer

The final volume of the balloon is approximately \( 238.77 \mathrm{mL} \).

Step 1: Convert the given temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, just add 273.15 to the Celsius temperature. Initial temperature in °C: \( 20.0 °C \) Initial temperature in K: \( 20.0 + 273.15 = 293.15 K \) Final temperature is given in Kelvin as: \( 1.00 \times 10^{2} K \)

Step 2: Write down the given values and the Charles' law formula

The given values are: Initial volume V1: \( 7.00 \times 10^{2} \mathrm{mL} \) Initial temperature T1: \( 293.15 K \) Final temperature T2: \( 1.00 \times 10^{2} K \) The Charles' law formula: \( V1/T1 = V2/T2 \)

Step 3: Substitute the given values into the Charles' law formula and solve for V2

Putting the values into the formula, we have: \( (7.00 \times 10^{2} \mathrm{mL})/293.15 K = V2 / 1.00 \times 10^{2} K \) Now, solve for V2: \( V2 = (7.00 \times 10^{2} \mathrm{mL}) \times (1.00 \times 10^{2} K / 293.15 K) \)

Step 4: Calculate the final volume

Perform the calculation for the final volume: \( V2 = (7.00 \times 10^{2} \mathrm{mL}) \times (1.00 \times 10^{2} K / 293.15 K) \) \( V2 = 700 \mathrm{mL} \times (100 K / 293.15 K) \) \( V2 = 700 \mathrm{mL} \times 0.3411 \) \( V2 = 238.77 \mathrm{mL} \) The final volume of the balloon is approximately \( 238.77 \mathrm{mL} \).

Key Concepts

Temperature Conversion

When dealing with gas laws and temperature calculations, converting temperatures between Celsius and Kelvin is essential. Kelvin is the preferred unit in scientific calculations because it starts from absolute zero, making it an 'absolute' scale. This is especially important in gas law calculations, such as Charles' Law.

• To convert Celsius to Kelvin, simply add 273.15 to the Celsius value.
• This adjustment accounts for the Kelvin scale starting point at absolute zero (-273.15°C).

In our problem, we converted the initial temperature from 20.0°C to 293.15K by adding 273.15. The final temperature was already given in Kelvin (100K), so no further conversion was needed. This ensures we are working with temperatures on the same scale, which is crucial when using formulas like Charles' Law that involve ratios of temperature and volume.

Gas Laws

Gas laws describe the behavior of gases in different conditions of volume, pressure, and temperature. Charles' Law is a fundamental principle in this field. It states that the volume of a gas is directly proportional to its temperature (in Kelvin) when the pressure is held constant.
Charles' Law can be expressed mathematically by the formula:

• \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]

This equation shows that if you know three of the variables, you can calculate the fourth one.
The key aspect of Charles' Law is maintaining constant pressure while adjusting for changes in temperature. It's important for understanding how gases expand or contract with temperature changes.

Volume Calculation

Determining the volume of a gas under different conditions involves applying Charles' Law and performing some calculations. Given initial and final temperatures, along with the initial volume, we want to find the final volume.
1. First, identify the given values:

• Initial Volume \(V_1 = 700 \text{ mL}\)
• Initial Temperature \(T_1 = 293.15 \text{ K}\)
• Final Temperature \(T_2 = 100 \text{ K}\)

2. Use the Charles' Law formula to find the final volume:
Substitute these values into the equation:

• \[\frac{700}{293.15} = \frac{V_2}{100}\]

3. Solve for \(V_2\):

• \[V_2 = 700 \times \frac{100}{293.15} \approx 238.77 \text{ mL}\]

This result of approximately 238.77 mL demonstrates how significantly the volume of a gas can decrease as its temperature is lowered. All calculations ensure that units are consistent and accurate analysis is possible.